DEBRE MARKOS UNIVERSITY

COLLEGE OF BUSINESS AND

ECONOMICS
DEPARTMENT OF MANAGEMENT (MBA)

Operation research assignment:-

Mekonnen Mehertu
ID No. GSE/398/12

Submitted to: Dr. Was

Wasihun Tiku, Ph.D
Bure, Ethiopia
June, 2020
 Question 1

An individual investor has Birr 70,000 to divide among several investments. The alternative investments
are municipal bonds with an 8.5% return, certificates of deposits with a 10% return, Treasury bill with a
6.5% return, and income bonds with a 13% return. The amount of time until maturity is the same for
each alternative. However, each investment alternative has a different perceived risk to the investor;
thus it is advisable to diversify. The investor wants to know how much to invest in each alternative in
order to maximize the return. The following guidelines have been established for diversifying the
investment and lessening the risk perceived by the investor.

1. No more than 20% of the total investment should be in an income bonds.

2. The amount invested in certificates of deposit should not exceed the amount invested in other three
alternatives.

3. At least 30% of the investment should be in treasury bills and certificates of deposits.

4. The ratio of the amount invested in municipal bonds to the amount invested in treasury bills should
not exceed one to three.

The investor wants to invest the entire Birr 70,000.

Required:

a. Formulate a LP model for the problem.

Solution
Decision variable
Four decision variables represent the monetary amount invested in each investment alternative
Let x1, x2, x3 and x4 represent present municipal bonds, certifi
certificate
cate of deposits. The Treasury bill
and income bonds respectively
Objective function
The problem is maximization because from the word problem we know that the objective of
the investor is to maximization return from the investment in the alternatives.
There for the objective function is expressed as
/max=0.085x1+0.1.x2+0.065x3+0.13x4
Where
 Z=total return from all investment
0.85 x1 return from the investment in municipal bonds
0.100x2 return from the investment in certificate in deposit
0.65 x3 return from the investment in treasury bills
x4 return from the investment in income bonds
Model constraint
In this problem constraints are the guide line established for diversifying the total investment.
Each guideline is transformed in to mathematical constraint separately
Guideline 1 above is transformed as
X4≤20%(70000)⇒x4≤14000
Guideline 2
Xn x1+x3+x4 ⇒x2-x1-x4≤0
Guideline 3
X2+x3≥30%(70000)⇒x2+x3≥21000
Guideline 4
X1.x3≤1/3 ⇒3x1≤x3⇒3x1-3x1-x3≤0
Finally
X1+x2+x+x4 = 70000, because the investor’s money equals to the sumof money invested in all
alternatives
This last constraint diff
differs
ers from ≤and ≥ inequalities previously developed in that there is a
specific requirement to invest as exact amount. There for,
for, the possibility to invest more or less
than birr 70000 is not considered. This problem contains all three types of constraints possible
in LP problem ≤,≥ and =as this problem demonstrates there is no restriction on mixing these
types of constraints
The complete LPM for this problem can be summarized as
Zmax=0.85 x1+ 0.1x2+0.065x3+0.13x4
Subjected to x4 ≤14000x2- x1-x3-x4≤0
X2+x3 ≥21000
3x1-x3≤0
X1+x2+x3+x4=70000
X1+x2+x3+x4 ≥0
Question 3

A sale manager has to assign salesmen to four territories. He has four candidates of varying
experience and capabilities and assesses the possible profit for each salesman in each territory
as given below. Find the assignment which maximizes profit?
Salesmen Territories

A B C D

1 35 27 28 37

2 28 34 29 40

3 35 24 32 33

4 24 32 25 82

Solution since it is a maximization problem we first subtract each of the entries in the table from the
latgest one which equals 82 here

Salesman Territory Raw mini

A B C D

1 47 55 54 45 45

2 54 48 53 42 42

3 47 58 50 49 47

4 58 50 57 0 0

Now we proceed as
Subtract value in each raw from every value in the raw. So the result will be
salesman Territory

A B C D

1 2 10 9 0

2 12 6 11 0

3 0 11 3 2

4 58 50 57 0

Column mini 0 6 3 0
Next subtract minimum value in each column in reduced cost from the above table then ew get

salesman Territories
A B C D

1 2 4 6 0
2 12 0 8 0

3 0 5 0 2
4 58 44 54 0

Since the number of aligned line raw and column are not we drew another ratable by finding unoccupied
column and from this column takes the list valued and ad to the intersection number and subtract from
unoccupied one. Finally we gate

salesman Territories

A B C D
1 2 4 0 0

2 12 0 2 0
3 0 11 0 8

4 58 44 48 0
Hence territories have been on zero elements C1,B2,A3 D4

Question 4

Peniel House and Furniture manufacturer produces two products: Beds and Chairs. Each unit of
Bed requires 3 hrs in molding unit, 4hrs in painting unit, and 1 hr in finishing. On the other hand,
each unit of Chair requires 3 hrs in molding unit, 2 hrs in the paint shop and 2 hours in finishing.
Each week, there are 210 hrs available in molding, 200hrs in painting, and 120 hrs in finishing
unit. The demand for Beds cannot exceed 40 units per week. Each unit of Bed contributes Birr
20 to profit, while each unit of chair contributes Birr 30. Determine the number of units of each
product per week to maximize the profit? (Use Graphic Method to solve the given LP problem).

Solution
Resource Beds Chair Maximum Available
Molding Unit 3 3 210
Painting Unit 4 2 200
Finishing Unit 1 2 120
Maximum Demand 40 Unlimited
Profit for each products 20 30
LP Model
Let X1 and X2 be the number of Beds and Chairs Respectively
Let the weekly profit be Z. Then. we must
MaxZ=20X1+30X2
Subject to
3X1+3X2≤210Molding Constraint
4X1+2X2≤200painting Constraint
X1+2X2≤120 Finishing Constraint
X1≤40 Demand Constraint
Plot the constraints (0.100) x2
3x1+3x2≤210 molding constraint (0.70)
3x1+3x2=210
If x1=0 D(0.60)
⇒3x1+3x2=210 C(80/3,140/3)
⇒3(0)+3x2=210
⇒x2=7 B(40,20)
The coordinates (0,70) FEASIBL x1+2x2≤120
If x2=0 REGION
⇒3x1+3(0)=210
⇒x1=70 (40,0) A (50,0) (70,0) (120,0) x1
The coordinates (70,0)
X1≤40
⇒painting constraint 4x1+2x2≤200 4x1+2x2≤200 3x1+3x2≤210
4x1+2x2=200
If x1=0
⇒x2=100, the coordinates (0,100) and if x2=0 ⇒x1=50, the coordinates (50,0)
⇒finishing constraint
X4+2x2≤120
X1+2x2=120 if x1=0⇒x2=60, the coordinates (60,0) And if
x2=0⇒x1=120,
x1=120, the coordinates (120,0)
1 2 3 4 5
corner Constraints Coordinates Objective function
points intersected X1 X2 20x1+30x2
A Demand 40 0 800
B Demand and painting 40 20 1400
C Painting and finishing 80/3 140/3 5800/3
D Finishing 0 60 1800
The maximum value of the objective function is 5800/3. Therefor the optimum solution is found
at the corner point C, where constraint painting and finishing are intersected. Hence optimal
solution x1=80/3, x2=140/3
Question 5

Construct the CPM network described by the following set of activities, compute the length of each path
in the network, and indicate the critical path.

Activity Time (months)

1–2 4

1–3 7

2–4 8

2–5 3

3–5 9

4–5 5

4–6 2

5–6 6

3–6 5

Answer
Construct the CPM network described by the following set of activities
4 3
2
8 5 5 6
1 4 2
7 3 9 6
5

Question 6
A firm plans to purchase at least 200 quintals of scrap containing high metal quality metal X and
low quality metal Y. It decides that the scrap to be purchased must contain at least 100 quintals
of X-metal and not more than 35 quintals of Y-metal. The firm can purchase the scrap from two
suppliers ( A and B) in unlimited quantities. The percentage of X and Y metals in terms of weight
in the scrap supplied by A and B is given below.
Metals Supplier A Supplier B

X 25% 75%

Y 10% 20%

The price of A’s scrap is Birr 200 per quintal and that of B is Birr 400 per quintal. The firm wants to
determine the quantities that it should buy from the two suppliers so that total cost is minimized? (Use
graphic approach)

Solution

LP Model
Let x3 and x2 be the number if high metal quality and low metal quality respecti
respectively.
vely.
Let the minimum cost be Z. then, we must
minZ=200x1+400x2
subject to
10%x1+20%x2≤35 quintal of metal y 25%
+75%x2≥100 quintal of metal B X1+x2≥200
quintals of metal Y and metal X

X1.x2≥0

(0,200)
X2
Plot the constraints
X1+x2≥200 quintals of metal Y&B (0,175)
X1+x2=200 C(50,150)
If x1=0
 Feasible region
⇒(0)+x2=200
⇒x2=200 (0,400/3 25% x1+75%x2 ≥100
The coordinates (0,200) B(100,100)
If x=0 A(250,50)
⇒x-3(0)0200⇒x=200 A(200,0) (350,0) (400,0) x1
The coordinates (200,0) L2
L3 L1

⇒ quintal if B constrait x1+x2 ≥200 10%x1+20%x2 ≤35

25%x1+75%x2 ≥100
25%x1+75%x2=100
If x1=0
⇒x2=400/3, the coordinates (0,400/3) and if x2=0 ⇒x1=400, the coordinates (400,0)
⇒quintal of Y constraint
10%x1+20%x2 ≤35
10%x1+20%x2=35 if =0 ⇒x2=175, the coordinates (0,175) and if x2=0, ⇒x1=350, the coordinates (50,0)
⇒ quintal of y constrait
10%x1+20%x2 ≤35
10%x1+20%x2 ≤35 if x1=0 ⇒ x2=175, the coordinates (0,175)
and if x2=0 ⇒ x1=350, the coordinates (350,0)

1 2 3 4 5
corner Constraints Coordinates Objective function
points intersected X1 X2 200x1+400x2
A L1 AND L3 325 25 75000
B L1 AND L3 100 100 60000
C L1 AND L2 50 150 70000
The minimum value of the object function is 60000. therefore, the optimum solution is found at the
corner point B , where constraint L1 and L3 are intersected. Hence, optimal solution: x1=100,x2=100.
Question 7
Six salesmen are to be allocated to six sales regions so that the cost of allocati
allocation
on of the job will be
minimum. Each salesman is cable of doing the job at different
different costs in each region. The cost (in birr)
matrix is given below.

Region Salesmen
I II III IV V VI

A 15 35 0 25 10 45
B 40 5 45 20 15 20

C 25 60 10 65 25 10
D 25 20 35 10 25 60
E 30 70 40 5 40 50
F 10 25 30 40 50 15

a. Find the allocation to give minimum cost. What is the minimum?

b. If the figures given in the above table represent the earning of each salesman at each
region, then find an allocation so that the earning will be maximum. Also workout this
maximum possible earning.

The minimum value of each row is subtracted from elments in the row and show reduced
table as follows as below
Region Salesmen

I II III IV V VI

A 25 35 0 25 10 45

B 35 0 40 15 10 15

C 25 50 0 55 15 0

D 25 10 25 15 0 50

E 25 65 35 0 35 45

F 0 15 20 30 40 5
 A. The minimum value of each column is subtracts from all elements in the columns an
d show reduced table as follows as below
Region Salesmen

I II III IV V VI

A 15 355 0 25 10 45

B 35 0 40 15 10 15

C 15 50 0 55 15 0

D 15 10 25 25 0 50

E 25 65 35 0 35 45

F 0 15 20 30 40 5

Since the number of lines covering all zeros 6 equal to the number of columns or rows, then the
allocation to give minimum cost are aiii, Bii,Ciii and Civ, Dv, Eiv, Fi
Then the minimum cost is 0+5+10+10+25+5+10=65
B.
Region Salesmen

I II III IV V VI

A 15 35 0 25 10 45

B 40 5 45 20 15 20

C 25 60 10 65 25 10

D 25 20 35 10 25 60

E 30 70 40 5 40 50

F 10 25 30 40 50 15

Since it is maximization problem we would first subtract each of the entries in the table
from the largest one which equals to 70. The result data are given below
Region Salesmen

I II III IV V VI
A 55 35 70 45 60 25

B 30 65 25 50 55 50
C 45 10 60 5 45 60

D 45 50 35 60 45 10
E 40 10 30 65 30 20
F 60 45 40 30 20 55
 We will use the minimum value process subtract the least value in each row and column.
The reduced table as follows.
Region Salesmen

I II III IV V VI
A 30 10 45 20 35 0

B 5 40 0 25 30 25
C 40 50 55 0 40 55

D 35 40 25 50 40 0
E 30 0 20 55 20 10

F 40 25 20 10 0 35

The least value of the columns are all zero then the reduced table is the same as the
above table
Since the number of lines covering all zeros is equal to the number of columns or rows,
then the allocation to give earning are AVI, BIII, CVI, DVI, EII, FV
The maximum possible earning is 45+20+65+60+70=310 birr
Then the minimum cost is 0+5+10+10+25+5+10=65
Question 14

The owner of the Burger Doodle Restaurant is considering two ways to expand operations:
opening a drive-up window or serving breakfast. The increase in profits resulting from these
proposed expansions depends on whether a competitor opens a franchise down the street. The
possible Profits from each expansion in operations given both future competitive situations are
shown in the following payoff table.
Competitor
Decision Open Not Open
Drive-up window $-6,000 $20,000
Breakfast 4,000 8,000
Select the best decision using the following decision criteria.

a. Maximax

b. Maximin

c. Equally likely

d. Minimax regret

solution

A. The decision maker selects the decision that will result in the maximum from each rows
.

Decision Competitor
Open Not open Maximum row

Drive-up window $-6,000 $20,000 $20,000

Breakfast 4,000 8,000 8,000

The maximum of maximum row is 20000. Then the best decision from the above table is
drive up window

B. The worst (minimum) payoff for each alternative

.

Decision Competitor

Open Not open Minimum row

Drive-up window $-6,000 $20,000 -6000

Breakfast 4,000 8,000 4000

From the above table the maximum of minimum row is 4000. Then the worst payoff
payoff for the alternatives
is 4000. There for the best decision criterion is breakfast

C. The best decision criterion

.

Decision Competitor
Open Not open average row

Drive-up window $-6,000 $20,000 70000

Breakfast 4,000 8,000 6000

 From the above table the maximum average row of the decision is 7000. There for the best
decision criterion is drive-up window

D..

Decision Competitor
Open Not open average row

Drive-up window $-6,000 $20,000 70000

Breakfast 4,000 8,000 6000

Identify the best payoff in column and then subtracting each of the other values in the column from that
payoff. That is

Decision Competitor
Open Not open Maximum row

Drive-up window -6000-4000=-10000 20000-20000=0 0

Breakfast 4000-4000=0 20000-8000=12000 12000

There for the minimum of maximum row is 0. Then the best decision is drive-up window

Question 15

Convert the following in to dual and solve it using simplex approach

Min Z=2x1 +8x2

Subject to:

5x1+ x2 > 10

2x1+ 2x2 > 14

X1+ 4x2 > 14

X1, x2 > 0

Solution

The dual of a normal max problem is a normal min problem and vice versa. Then, yje dual of minimum is
the maximum. There for yhr dual of
Min Z=2x1+8x2 subject to; 5x1+x2≥ 10

2x1+2x2≥ 14

X1+4x2≥ 14

X1,x2≥ 0

Can be calculated by interchangeable the column of min into rows of max

Max Z=10x1+14x2+14x3 subject to:5x1+2x2+x3≤ 2

X1+2x2+4x3≤ 8

X1.x2.x3≥ 0

Change into standard dorm of the above inequality

Max Z=10x2+14x2+14x3+0s1+0s2 subject to ;5x1+2x2+x3+s1=2

X1+2x2+4x3+s2=8

X1.x2.x3≥ 0

Initial feasible solution if x1=0,x2=0,x3=0 then s2=2 and ifx1=0,x2=0,x3=0 then s2=8

The initial simplex table is

Basic Cj 10 14 14 0 0 quantity
variable X1 X2 X3 S1 S2
S1 0 5 2 1 1 0 2/2=1

S2 0 1 2 4 0 1 8/2=4
Zj 0 0 0 0 0 0

Cj-zj 10 14 11 0 0 0
Pivot element pivot column pivot row

The row cj-zj the largest value is 14 at the column of x2 and x3. Then we take at a column of x2 and the
smallest rati
ratio
o1

Table
Table 1
Divided all element of R1 by 2 to change pivot element into one and R2=R2-2R1 we get the following
table

Basic Cj 10 14 14 0 0 quantity
variable
X1 X2 X3 S1 S2
X2 14 5/2 1 1/2 ½ 0 1/1/2=2

S2 0 -4 0 3 -1 1 6/3=2
Zj 35 14 7 7 0 14

Cj-zj 25 0 7 -7 0 14
pivot column Pivot element pivot row

Divide all elements of pivot row by 3 to change pivot element into one and R1=R1-1/2R2 then, we get
the following table

Basic Cj 10 14 14 0 0 quantity
variable
X1 X2 X3 S1 S2

X2 14 19/6 1 0 2/3 -1/6 0

X3 14 -4/3 0 3 -1/3 1/3 2

Zj 77/3 14 14 14/3 7/3

Cj-zj -47/3 0 0 -14/3 -7/3 28

From the above table the cj-zj row elements are negati
negative
ve and zero then, minimum value at x1=0, x2=0
and x3=2 is 28

Question 18

An investor is considering investing in stock, real estate, or bonds under uncertain economic
conditions. The payoff table of returns for the investor’s decision situation is shown below
Economic Conditions
Investment Good Stable Poor

Stocks $ 5,000 $ 7,000 $ 3,000

Real estate -2,000 10,000 6,000

Bonds 4,000 4,000 4,000

a. What is Equal likely decision?

b. What is the Maxim in decision?

c. what is the Maxi max decision?

d. what is the Hurwitz decision ? Use = 0.3

e. What is the Mini max regret decision?

Solution

Investment Economic condition

Good Stable Poor Average row

Stocks 5000 7000 3000 5000

Real estate -2000 10000 6000 4700

bonds 4000 4000 4000 4000

The maximum of average row for the above table is 5000. There for stocks are the best decision

B. the maximum van be calculated from the given alternatives be finding maximum of
elements of each row from minimum of each element

Investment Economic condition

Good Stable Poor Minimum row

Stocks 5000 7000 3000 3000

Real estate -2000 10000 6000 -2000
bonds 4000 4000 4000 4000

From the minimum row column the maximum value is 4000, then the investment decided the best
decision is bond

C. the maximum decision states that the maximum value of each row

Investment Economic condition

Good Stable Poor Maximum row

Stocks 5000 7000 3000 7000

Real estate -2000 10000 6000 10000

bonds 4000 4000 4000 4000

The maximum of maximum row from the above table is 10000. Thre for the best decision of the
investment is real state

D.

Investment Economic condition

Good Stable Poor maximum row Minimum row

Stocks 5000 7000 3000 7000 3000

Real estate -2000 10000 6000 10000 -2000

Bonds 4000 4000 4000 4000 4000

If a=0.3 then the optimism is 0.3 and 0.7 is pessimism, there for Hurwitz decision for each
investment can be calculated as follows

Stocks=0.3x7000+0.7x3000=4200

Real state= 0.3x1000+0.7x-2000=1600

Bonds =0.3x4000+4000x0.7=4000

The maximum value of the result is 4200, then the best decision of the investment is stocks.

Investment Economic condition

 Good Stable Poor

Stocks 5000 7000 3000

Real estate -2000 10000 6000

bonds 4000 4000 4000

To calculate maximum regret first find maximum value for each column of the given alternatives and
then subtract each elements of the column from the maximum value of each column

Second find the maximum value of each row and then finally find minimum of each maximum row. That
means the maximum value of each column respectively are 5000,10000,6000

The diff
difference
erence between the maximum value and each elements of columns are as follows

5000-5000=0, 5000 –(-2000)=7000,5000 – 4000=1000

10000 – 7000=3000, 10000 – 10000=0, 10000 – 4000=6000

6000 – 3000=3000, 6000 -6000=0, 6000 – 4000=2000

From the above calculation the decision table as follows

Investment Economic condition

Good Stable Poor Maximum row

Stocks 0 3000 3000 3000

Real estate 7000 0 0 7000

bonds 1000 6000 2000 6000

From the above table the minimum value of maximum row is 3000. There for the opportunity loss the
given alternatives is stocks

Question 26

A large steel manufacturing company has three options with regard to production. (i) Produce
commercially (ii) build pilot plant (iii) stop producing steel. The management has estimated that
their pilot plant, if built, has 0.8 chance of high yield and 0.2 chance of low-yield . if the pilot
plant does show a high yield , management assigns a probability of 0.75 that the commercial
plant will also have a high yield. If the pilot plant shows a low yield, there is only a 0.1 chance
that the commercial plant will show a high yield. Finally management best assessment of the
yield on a commercial-size plant without building a pilot plant first has a 0.6 chance of high
yield. A pilot plant will cost birr 300,000. The profit earned under high and low yield conditions
are birr 12000,000 and birr 1200,000 respectively. Find the optimum decision for the company?
Answer
EMV at Decision Node 1 = max (75,96,000, 76,80,000, 0)
Question 27

What is the role of decision making in operation research? Define scientific decision-making and
explain it affects
ffects operation research decisions.
Answer
Diff
Different
erent methods such as stimulation, queuing theory, game theory, mathematical &
network analysis which are part of operati
operation
on research are considered as major helpers
in the decision-
decision-making process. ... Effective
ffective Decisions Operati
Operation
on Research helps the
managers to make well-informed and effective
ffective decisions.
decisions.

scientific decision making.

making. Systematic approach to collecting facts and applying
logical decision making techniques, instead of generalizing from experience, intuition
(guessing), or trial and erro
error.
Question 28

Explain how and why operation research methods have been valuable in aiding executive
decision?
 Operation research is an aid for the executive in making his/her decision bt
providing the needed quantitative information, based on scientific method
analysis.
Question 29

Explain approximation in operation research model. What are advantage and characteristics of a
good model?
 A model in the sense used in OR is defined as representation of an actual or
situation
  A model is constructed to analyze and understand the give system for the
purpose of improving its performance. The reliability of the solution obtained
from a model depends on the validity of the model in representing the system
under study. A model, allows the opportunity to examine the behavioral changes
of a system without disturbing the on-going operations
Advantages of model
 Models in general are used as an aid for analyzing complex problem. However a
model can also serve other purposes as
Question 30

26. What are the situations where operation research techniques will be applicable?
Answer
Today,
Today, almost all fields of business and government utilizing the benefits of Operations
Research. There are voluminous of applications of Operations Research. Although it is
not feasible to cover all applications of O.R. in brie
brief. The following are the abbreviated
set of typical operations research applications to show how widely these techniques are
used today:
 Accounting: Assigning audit teams effectively
ffectively Credit policy analysis Cash flow
planning Developing standard costs Establishing costs for by products Planning of
delinquent account strategy
 Construction: Project scheduling, monitoring and control Determination of proper
work force Deployment of work force Allocation of resources to projects Facilities
 Planning: Factory location and size decision Estimation of number of facilities
required Hospital planning Quantitative Techniques for Managers International
logistic system design Transportation loading and unloading Warehouse location
decision
 Finance: Building cash management models Allocating capital among various
alternatives Building financial planning models Investment analysis Portf
Portfolio
olio analysis
Dividend policy making Manufacturing: Inventory control Marketing balance
projection Production scheduling Production smoothing
  Marketing: Advertising budget allocation Product introduction timing Selection of
Product mix Deciding most effective
ffective packaging alternative Organizational Behavior /
 Human Resources: Personnel planning Recruitment of employees Skill balancing
Training program scheduling Designing organizational structure more effectively
ffectively
 Purchasing: Optimal buying Optimal reordering Materials transfer Research and
Question 31 For what type of business problems might game theory helpful?

The application of game theory is not limited to games in ordinary sense

of it but also includes in economics, warfare and social behavior, etc. To
analyze the theory of games we introduce the terms player, pay off and
saddle point.

In business management, economics and commerce we come across competitive

situations involving conflicting interests. To tackle such situations a special discipline
called game theory has been developed. In game theory we consider two more persons
with different objectives, each of whose actions influence the outcomes of the other.
Each player wants to maximize his minimum expected gain or minimize his maximum
expected loss.
The aim of the theory of games is to analyze the different situations each player has to
face and different situation he has to choose according to those of the opponent, The
term game represents a conflict between two or more parties. Some of the recreational
games such as checkers, chess, bridge, etc., can be analyzed as games of strategy. But
games such as dice and roulette are not games of strategy since a person playing one of
these games is playing against the odds and not against an opponent.
Games are classified according to the number of distinct objectives present in the game.
The number of persons present in the game is not necessarily the same as the number
of people playing the game. For example, if two or more persons form a group in which
they agree to share their gains, they treated as single person.

A player is called a competitor. A strategy for a player is a plan which specifies his action
for every possible action of his opponent. In a two person game if player I has 'm'
strategies and player II has 'n' strategies then the game is said to be a m x n game. If the
number of strategies of the players is finite then the game is said to be a finite game. If
at least one of the players has in finite number of strategies, the game is said to be
infinite.
A game with two players wherein one person's gain is the loss of the other is call two
person zero sum game. The pay offs corresponding to various strategies of the players
are presented in a matrix called the game matrix.